If int frac{1}{xleft[(log x)^2+4 log x-1right | vedclass

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(A) Let $I = \int \frac{1}{x[(\log x)^2 + 4 \log x - 1]} dx$. Substitute $\log x = t$,so $\frac{1}{x} dx = dt$. The integral becomes $I = \int \frac{1

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$A=\frac{1}{2 \sqrt{5}}, B=(2-\sqrt{5}), C=(2+\sqrt{5})$

Vedclass · Answer

(A) Let $I = \int \frac{1}{x[(\log x)^2 + 4 \log x - 1]} dx$. Substitute $\log x = t$,so $\frac{1}{x} dx = dt$. The integral becomes $I = \int \frac{1}{t^2 + 4t - 1} dt$. Complete the square in the denominator: $t^2 + 4t - 1 = (t+2)^2 - 5 = (t+2)^2 - (\sqrt{5})^2$. Using the standard integral formula $\int \frac{1}{u^2 - a^2} du = \frac{1}{2a} \log \left| \frac{u-a}{u+a} \right| + K$,we get: $I = \frac{1}{2\sqrt{5}} \log \left| \frac{(t+2) - \sqrt{5}}{(t+2) + \sqrt{5}} \right| + K$. Substituting $t = \log x$ back: $I = \frac{1}{2\sqrt{5}} \log \left| \frac{\log x + 2 - \sqrt{5}}{\log x + 2 + \sqrt{5}} \right| + K$. Comparing this with the given form,we find $A = \frac{1}{2\sqrt{5}}$,$B = 2 - \sqrt{5}$,and $C = 2 + \sqrt{5}$.

If $\int \frac{1}{x\left[(\log x)^2+4 \log x-1\right]} d x=A \log \left[\frac{\log x+B}{\log x+C}\right]+K$ where $K$ is the constant of integration,then

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